Mixtures & Alligation

🔒 Log in to track

high importance~1 Q in Tier 120 formulas⚡ 15 shortcuts5 subtopics

Overview Learn (5)Formulas (20)Examples (5)Practice (66)

All subtopics·Subtopic 1 of 5

Mixture Concentration & Amounts

🔒 Log in to track

⏱ 5 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

A mixture holds two or more ingredients. Two quantities run every question:

Amount of each ingredient, from the ratio: in 40 L of milk : water = 3 : 1, milk is $40 \times \dfrac{3}{4} = 30$ L.
Concentration of one ingredient: amount ÷ total volume, as a per cent.

When something is added or removed, track the amounts. Adding water grows only the water. Removing mixture takes both ingredients in proportion.

What a mixture is

A mixture is two or more ingredients put together — milk and water, rice at two prices, copper and zinc in an alloy.

Every question turns on two quantities:

The amount of each ingredient (litres, kg, grams).
The concentration of one ingredient: its share of the total, as a per cent.

Rule: Track ingredient amounts, not the ratio alone. Ratios move when only one column changes.

Mean price of a blend

Mix items at different prices and the blend has one average price.

\text{mean price} = \frac{\text{total value}}{\text{total quantity}} = \frac{q_1 p_1 + q_2 p_2}{q_1 + q_2}

Example: 40 kg of rice at ₹6 mixed with 60 kg at ₹7. Value = $240 + 420 = ₹660$ over 100 kg. Mean = ₹6.60 per kg.

This one line answers every "find the average price of the mixture" question.

From ratio to real amounts

A ratio such as 7 : 5 splits the total into 12 parts.

One part = total ÷ sum of ratio terms.
Each ingredient = its share × one part.

Example: A 72-litre mixture has milk and water in the ratio 7 : 5. One part = $72 \div 12 = 6$ L. Milk = $7 \times 6 = 42$ L, water = 30 L.

Adding one ingredient

Adding water changes only the water column — the milk stays exactly as it was. That fixed ingredient is the anchor of the question.

Example: A 40 L mixture has milk : water = 3 : 1. How much milk makes it 4 : 1? Water stays 10 L. Milk must reach $4 \times 10 = 40$ L. Add $40 - 30 = 10$ L.

Routine: write the old amounts, freeze the ingredient not being added, then solve for the addition.

Percentage strength

Concentration works the same way in per cent. A 150 L solution at 60% acid holds $90$ L of acid.

Add x litres of pure acid (strength 100%) and the total grows too:

\frac{90 + x}{150 + x} = \frac{3}{4} \Rightarrow x = 90 \text{ L}

Watch: The denominator grows with every addition. A per cent is always a share of the current total.

Per cent of the mixture vs per cent of the other item

Read per cent questions twice. "Water is 25% of the mixture" means water : milk = 25 : 75 = 1 : 3 — not 1 : 4.

Convert the other way just as carefully: milk : water = 7 : 5 makes milk $\dfrac{7}{12} = 58\frac{1}{3}\%$ of the 72 L mixture. The sum of ratio terms is the denominator, never the other ingredient.

Careful: Exam options are built to catch the wrong denominator. Write the fraction, then match the option.

Alloys: the same arithmetic

An alloy is a mixture in grams. Copper : zinc = 5 : 3 in a 24 g piece gives 15 g copper and 9 g zinc. To make it 3 : 1, freeze zinc at 9 g, push copper to 27 g, add 12 g.

Tip: Prices, litres, grams — the units change, the method never does.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Mean price of a mixture

How to spot it:

Quantities and unit prices are given; the average price of the blend is asked.

\text{mean} = \frac{\sum q_i p_i}{\sum q_i}

Method

Multiply each quantity by its price and add: total value.
Add the quantities: total weight or volume.
Divide. A decimal like ₹6.60 is a clean exam answer.

Why it works:

The mean value is weighted by quantity, never a plain middle of the prices.

Try this

40 kg of rice at ₹6 per kg is mixed with 60 kg of rice at ₹7 per kg. Find the average price of the mixture.

Show solution

Value = $40 \times 6 + 60 \times 7 = 240 + 420 = 660$ .
Quantity = $100$ kg.
Mean = $660 \div 100 = ₹6.60$ per kg.

Answer

₹6.60 per kg

Type 2very common2 practice Q

Ratio to absolute amounts

How to spot it:

A ratio (milk : water 7 : 5) with a total quantity is given — one part is asked.

\text{one part} = \frac{\text{total}}{\text{sum of ratio terms}}

Method

Divide the total by the sum of the ratio terms.
Multiply each term by one part.
Remember which operations keep a column fixed.

Why it works:

A ratio is relative; the total converts it into litres or kg.

Try this

A 72-litre mixture has milk and water in the ratio 7 : 5. Find the quantity of water in it.

Show solution

Parts = $7 + 5 = 12$ ; one part = $72 \div 12 = 6$ L.
Water = $5 \times 6 = 30$ L (milk = 42 L).

Answer

30 litres

Type 3very common2 practice Q

Adding an ingredient to change the ratio

How to spot it:

'How much X must be added so the ratio becomes a : b?'

\frac{\text{component}_1}{\text{component}_2 + x} = \frac{a}{b} \ \text{or}\ \frac{\text{component}_1 + x}{\text{component}_2} = \frac{a}{b}

Method

Get the old amounts from the ratio and the total.
Freeze the ingredient not being added.
Set the new ratio as an equation and solve for x.

Why it works:

Only the added column grows; the other ingredient is untouched.

Try this

A 40-litre mixture has milk and water in the ratio 3 : 1. How much milk must be added to make the ratio 4 : 1?

Show solution

Old amounts: milk 30 L, water 10 L.
Water stays 10 L; at 4 : 1 milk must be $4 \times 10 = 40$ L.
Add $40 - 30 = 10$ L.

Answer

10 litres

Type 4common2 practice Q

Percentage strength of a solution

How to spot it:

A solution's concentration is given; after adding solvent or pure solute, the new strength is asked.

\text{strength} = \frac{\text{solute}}{\text{total volume}} \times 100

Method

Compute the solute amount from the percentage.
Add to the solute (or leave it) and grow the total as the story says.
Write the new fraction and convert to a per cent.

Why it works:

Percentages are fractions of the current total, and the total changes with every addition.

Try this

150 L of a solution contains 60% acid. How much pure acid must be added to make it 75% acid?

Show solution

Acid = $90$ L. Let the addition be $x$ .
$\frac{90 + x}{150 + x} = \frac{3}{4} \Rightarrow 360 + 4x = 450 + 3x$ .
$x = 90$ L.

Answer

90 litres

Type 5common2 practice Q

Alloys and blending by parts

How to spot it:

Alloys with metal ratios, or blends whose parts are used again in another mix.

\text{component} = \frac{\text{ratio term}}{\text{sum}} \times \text{total}

Method

Convert each alloy's ratio into grams (or kg) of each metal.
Pool the columns across all sources.
Read the new ratio, or solve for the addition needed.

Why it works:

Metals from different sources simply add up, column by column.

Try this

An alloy has copper and zinc in the ratio 5 : 3. In 24 g of it, how much copper must be added to make the ratio 3 : 1?

Show solution

24 g split 5 : 3 → copper 15 g, zinc 9 g.
Zinc stays 9 g; at 3 : 1 copper must be $27$ g.
Add $27 - 15 = 12$ g.

Answer

12 g

Formula sheet

Amount from ratio

\text{part} = \text{total} \times \frac{\text{share}}{\text{sum of shares}}

Concentration

C = \frac{\text{ingredient}}{\text{mixture}} \times 100\%

Mean price

\text{mean} = \frac{q_1 p_1 + q_2 p_2}{q_1 + q_2}

Adding water

C' = \frac{A}{M + w}

A = amount of the other ingredient, unchanged.

Removing mixture

\text{each ingredient shrinks in its own share}

A uniform draw keeps the ratio.

Shortcuts that save time

⚡ Freeze the unchanged ingredient

Water added means milk unchanged. Hang the whole question on the ingredient that does not move.

Example

A 40-litre mixture has milk and water in the ratio 3 : 1. How much water does it contain?

Show solution

Parts = $3 + 1 = 4$ ; one part = $40 \div 4 = 10$ L.
Water = $1 \times 10 = 10$ L.

Answer

10 litres

⚡ Rebuild the per cent after dilution

Milk is fixed, the total grows — divide again.

Example

20 litres of a mixture contains 60% milk. After adding 5 litres of water, find the milk percentage.

Show solution

Milk = $60\%$ of 20 = 12 L, unchanged.
New total = $25$ L.
Milk % = $12 \div 25 = 48\%$ .

Answer

48%

⚡ One ratio change, one equation

Set the frozen ingredient against the wanted ratio and solve for the addition.

Example

A 50-litre mixture of milk and water is in the ratio 4 : 1. How much water must be added to make it 2 : 1?

Show solution

Milk = $40$ L stays; water = 10 L.
Target 2 : 1 needs water = $40 \div 2 = 20$ L.
Add $20 - 10 = 10$ L.

Answer

10 litres

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Changing both parts of a ratio when only one moved.

Adding water leaves the milk amount untouched. Adjust only the water and the total.

Mistake 02

Thinking x litres of water cuts the milk percentage by x%.

Recompute: milk ÷ new total. The drop depends on the sizes.

Mistake 03

Taking the share of the wrong ingredient (1/4 instead of 3/4).

Label the ratio first: milk : water = 3 : 1 means milk takes 3 of the 4 parts.

Mistake 04

Forgetting the total volume changes when anything is added.

Every addition grows the denominator of the concentration.

Mistake 05

Reading '25% water' as water : milk = 25 : 100.

Per cent is of the mixture: water : milk = 25 : 75 = 1 : 3.

Quick revision

Read this the night before the exam.

Mean price = total value ÷ total quantity.
One part = total ÷ sum of ratio terms.
Adding X moves only X's column and the total.
New ratio → one equation in the one added amount.
Per cent of mixture ≠ per cent of the other ingredient.

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.

Continue learning